Terence Tao real analysis Chap5- proposition 5.4.14

Proposition 5.4.14 Given any two real numbers \ (X <Y \) , we can find a rational \ (Q \) such that \ (X <Q <Y \) .
Proof:
set \ ({X = n-LIM_ \ to \ infty A_N}, Y = {n-LIM_ \ to \ infty} B_n \)
\ (\ Because X <Y \) , (sorted real numbers) defined by 5.4.6 defined 5.4.3 (positive and negative real numbers) to give
\ (yx = LIM_ {n \ to \ infty} (b_n-a_n) \) is positive \ (= LIM_ {n \ to
\ infty} c_n \) where \ ( {(c_n) ^ {\ infty } _ {n = 1}} \) is moving away from the \ (0 \) Cauchy sequence, $ {\ forall} n \ geq 1, c_n \ geq c> 0, c \ in \ mathbb {Q} $
defined by 5.3.1 (equal to the real numbers) to give \ ((b_n-a_n) ^ {\ infty} _ {n = 1} \) and \ ((c_n) ^ {\ infty} _ {n = 1} \) equivalent, i.e.
\ ({\ forall} \ varepsilon_1 > 0, {\ exists} N_1 \ in \ mathbb {N}, st \ vert b_n-a_n-c_n \ vert \ leq \ varepsilon_1 \ ) to\ ({\ forall} n \ geq N_1 \) established
defined by 5.1.8 (Cauchy sequence) to give
\ ({\ forall} \ varepsilon_2 > 0, {\ exists} N_2 \ in \ mathbb {N}, st \ vert a_n-a_ {N_2} \ vert \ leq \ varepsilon_2 \) of \ ({\ forall} n \ geq N_2 \) established
\ ({\ forall} \ varepsilon_3 > 0, {\ exists} N_3 \ in \ mathbb { N}, st \ vert b_n- b_ {N_3} \ vert \ leq \ varepsilon_3 \) of \ ({\ forall} n \ geq N_3 \) established
order \ (\ varepsilon_1 = \ varepsilon_2 = \ varepsilon_3 = \ frac {c } {. 4} \) , \ (N = max \ {N_l, N_2, N_3 \} \) , for \ ({\ FORALL} n-\ GEQ N \)
\ (\ Vert B_n-A_N-C_N \ Vert \ Leq \ frac {c} {4} \ Rightarrow c- \ frac {c} {4} = \ frac {3c} {4} \ leq c_n- \ frac {c} {4} \ leq b_n-a_n \ leq c_n + \ frac {c} {4} \ Rightarrow \ frac {3c} {4} \ leq c_N- \ frac {c} {4} \ leq b_N-a_N \ leq c_n + \ frac {c} {4} \)
\(\vert b_n-b_N\vert \leq \frac{c}{4}\Rightarrow b_N-\frac{c}{4} \leq b_n \leq b_N+\frac{c}{4}\)
\(\vert a_n-a_N\vert \leq \frac{c}{4}\Rightarrow a_N-\frac{c}{4} \leq a_n \leq a_N+\frac{c}{4}\)
取\(q=\frac{a_N+b_N}{2}\)
\(b_n-q=b_n-\frac{a_N+b_N}{2} \geq b_N-\frac{c}{4}-\frac{a_N+b_N}{2}=\frac{b_N-a_N}{2}-\frac{c}{4} \geq \frac{3c}{8}-\frac{c}{4}=\frac{c}{8}>0\)，\(n \geq N\)
\(q-a_n=\frac{a_N+b_N}{2}-a_n \geq \frac{a_N+b_N}{2}-a_N-\frac{c}{4}=\frac{b_N-a_N}{2}-\frac{c}{4} \geq \frac{3c}{8}-\frac{c}{4}=\frac{c}{8}>0\)，\(n \geq N\)
令\((D_n)^{\infty}_{n=1}=\begin{equation} \left\{ \begin{array}{c} \frac{c}{8}，n<N \\ b_n，n \geq N \\ \end{array} \right. \end{equation}\)与\ ((b_n-q) ^ {\ infty} _ {n = 1} \) equivalence, \ (n-YQ = LIM_ {\ to \ infty} (B_n-Q) = {n-LIM_ \ to \ infty} D_n \)
\ (\ Because {\ FORALL} n-\ GEQ. 1, D_n \ GEQ \ FRAC {C} {. 8} \) , \ (\ THEREFORE (D) ^ {\ infty} _ {n-=. 1} \) is moving away \ (0 \) Cauchy sequence, \ (n-YQ = LIM_ {\ to \ infty D_n} \) is positive, i.e. \ (y> q \)
Similarly order \ ((E_n) ^ {\ infty } _ {n = 1} = \ begin {equation} \ left \ {\ begin {array} {c} \ frac {c} {8}, n <N \\ a_n, n \ geq N \\ \ end { array} \ right. \ end { equation} \) and \ ((q-a_n) ^ {\ infty} _ {n = 1} \) equivalence, \ (n-QX = LIM_ {\ to \ infty} (Q -a_n) = LIM_ {n \ to
\ infty} E_n \) and \ (\ because {\ forall} n \ geq 1, E_n \ geq \ frac {c} {8} (E_n) ^ {\ infty} _ { n = 1} \) is moving away from the Cauchy sequence 0, \ (\ {n-LIM_ THEREFORE QX = \ to \ infty of E_n} \)Is positive, i.e. \ (q> x \)
In summary, for any two real numbers \ (X <Y \) , there is always a rational number \ (Q \) such that \ (x <q <y \ )